判定级数ntan (π2∧n+1)的敛散性

题目

判定级数ntan (π2∧n+1)的敛散性
如题 判定级数 ntan (π2∧n+1）的敛散性 括号里是π 除以2的n+1次方

答案

答：
limn->∞ u(n+1)/u(n)
=limn->∞ [(n+1)tan(π/2^(n+2))]/[ntan(π/2^(n+1))]
又当t->0时,tant~t
=limn->∞ [(n+1)(π/2^(n+2))]/[n(π/2^(n+1))]
=limn->∞ (n+1)/(n*2)
=1/2